Perpendicular form of the straight line equation. There are $5$ to $6$ standard forms of the straight line equation. for example slope intercept form, two intercept form, point slope form and perpendicular form. I have clear visualization of all forms except the perpendicular form. 
Can any one help me out to give me the concept of perpendicular form. and let me know that how to reduce an equation of straight line for example $x-2y-3=0$ into perpendicular form, i.e. $$x\cos\alpha+y\sin\beta=p$$
 A: Using The Pearson Complete Guide For Aieee 2/e
 By Khattar as a point of reference for the items below.
You want to convert 
$$\tag 1 x-2y-3=0$$
into normal (perpendicular) form.
We have:
$$Ax + By = -C \rightarrow (1)x + (-2)y = -(-3)$$
This means $C \lt 0$, so we divide both sides of $(1)$ by:
$$\sqrt{A^2 + B^2} = \sqrt{(1)^2 + (-2)^2} = \sqrt{5}$$
This yields:
$$\dfrac{1}{\sqrt{5}} x - \dfrac{2}{\sqrt{5}} y = -\dfrac{-3}{\sqrt{5}}$$
In normal form, this is:
$$\cos(\alpha)~x + \sin(\alpha)y = p \rightarrow \dfrac{1}{\sqrt{5}} x - \dfrac{2}{\sqrt{5}} y = -\dfrac{-3}{\sqrt{5}}$$
Note that this has a positive $x$ coordinate, and a negative $y$ coordinate, which puts us in the $4^{th}$ quadrant.
So, we have:
$$\cos(\alpha) = \dfrac{1}{\sqrt{5}}, \sin(-\alpha) = -\dfrac{2}{\sqrt{5}} , p = -\dfrac{-3}{\sqrt{5}} = \dfrac{3}{\sqrt{5}}$$
This gives us:
$$\alpha \approx 1.10714871779409 ~\mbox{radians}~ \approx 63.434948822922 {}^{\circ}$$
Since we are in the $4^{th}$ quadrant, we can write this angle as:
$$\alpha \approx 2 \pi - 1.10714871779409 ~\mbox{radians}~ \approx 5.176036589385497 ~\mbox{radians}~ \approx 296.565 {}^{\circ}$$
As another reference point, see $10.3.4$ (including examples) at Straight Lines.
A: I felt exactly same when first studying the perpendicular form. Seeing components of $p\ \mathrm{as}\ x\cos(\alpha)+y\sin(\alpha)$ line segments by means of a simple construction could burn it into your visual memory:
$$p = OL = OQ + QL = OQ + RP =x\cos(\alpha)+y\sin(\alpha)$$

